Get list of character which not satisfy Like

Question

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SQL-Server-2008

I need list of character that not satisfy like condition for example,

--Validation for X
declare @test varchar(50)
set @test='ds[fds'
if @test like '%[^0-9a-zA-Z .()/-]%'
print 'invalid -yes'
else
print 'invalid -no'

In above case i need output '[' which is not satisfy condition also if there is more than one invalid character then need that all

SQL

declare @phone varchar(50)
set @phone='212@@f'

select  REPLACE(@phone
               , SUBSTRING(@phone, PATINDEX('%[^0-9a-zA-Z .()/-]%', @phone), 1)
              , '')

but it does not work if there is sql character like '%' in input string so what is the solution.

SQL

declare @phone varchar(50)
set @phone='212%@@f'

select  REPLACE(@phone
               , SUBSTRING(@phone, PATINDEX('%[^0-9a-zA-Z .()/-]%', @phone), 1)
              , '')

Posted 20-Mar-14 2:19am

vishal_h

Updated 20-Mar-14 17:45pm

v3

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Sergey Vaselenko · Answer 1 · 2014-03-20T22:25:00

SQL

declare @phone varchar(50)
set @phone='212%@@f'

declare @illegal varchar(50)
declare @test varchar(50)

set @illegal = SUBSTRING(@phone, PATINDEX('%[^0-9a-zA-Z .()/-]%', @phone), 1)
set @test = REPLACE(@phone, SUBSTRING(@phone, PATINDEX('%[^0-9a-zA-Z .()/-]%', @phone), 1), '')

while @test <> @phone
    begin
    set @phone = @test
    set @illegal = @illegal + SUBSTRING(@phone, PATINDEX('%[^0-9a-zA-Z .()/-]%', @phone), 1)
    set @test = REPLACE(@phone, SUBSTRING(@phone, PATINDEX('%[^0-9a-zA-Z .()/-]%', @phone), 1), '')
    end

select @illegal
select @phone